(a) Explanation Given Information: The mass ( m ) of the sheet steel is 2.4 kg. The dimension of the sheet steel is 160mm X 640 mm. The component is at rest when a couple ( M 0 ) for is ( 0.8 N ⋅ m ) k is applied to it. Calculation: The system is split in three sections consisting of two square plates and a rectangular plate to calculate the area. Calculate the area ( A ) of the system using the relation: A = b 2 + ( 2 b ) b + b 2 Here, side is b . Substitute 160 mm for b . A = ( 160 mm ) 2 + ( 2 × 160 mm ) ( 160 mm ) + ( 160 mm ) 2 = 25 , 600 + 51 , 200 + 25 , 600 = ( 102 , 400 mm 2 ) ( m 2 10 6 mm 2 ) = 0.1024 m 2 Calculate the mass per unit area of the steel using the relation: ρ = m A Here, ρ is mass per unit area. Substitute 2.4 kg for m and 0.1024 m 2 for A . ρ = 2.4 kg 0.1024 m 2 = 23.4375 kg / m 2 Express the angular moment ( H G ) of the system about the mass centre. H G = − I x z ω i − I y z ω j + I z ω k Here, I x z is mass moment of inertia about xz axis, I y z is mass moment of inertia about yz axis, I z is mass moment of inertia about the z axis, and ω is angular velocity. The moment of inertia about the xz plane is zero. H G = − I y z ω j + I z ω k (1) Express the angular momentum ( H ˙ G ) G x y z of the reference frame Gxyz. ( H ˙ G ) G x y z = − I x z α i − I y z α j + I z α k Express the angular velocity of the reference frame Gxyz . Ω = ω k (2) Here, Ω is angular velocity of reference frame is. The moment of inertia ( H ˙ G ) G x y z about the xz plane is zero. ( H ˙ G ) G x y z = − I y z α j + I z α k (3) Express the rate of change of angular momentum. H ˙ G = ( H ˙ G ) G x y z + Ω × H G (4) Substitute Equation (1), (2), and (3) in Equation (4). H ˙ G = ( − I y z α j + I z α k ) + ω k × ( − I y z ω j + I z ω k ) The matrix multiplication for vector product is done. H ˙ G = − I y z α j + I z α k + I y z ω 2 i (5) Express the moment of inertia ( I z ) a r e a about the z axis. ( I z ) a r e a = plate A + plate A B + plate B = [ 1 6 b 4 + b 2 ( 1 2 b ) 2 ] + [ 1 12 ( 2 b ) b 3 ] + [ 1 6 b 4 + b 2 ( 1 2 b ) 2 ] = 1 6 b 4 + 1 4 b 4 + 1 6 b 4 + 1 6 b 4 + 1 4 b 4 = b 4 Express the moment of inertia ( I y z ) a r e a about the yz axis (b) To determine The dynamic reaction at A ( A) and B ( B ) immediately after the couple is applied.

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics

11th Edition

ISBN: 9781259639272

Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self

Publisher: McGraw-Hill Education

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Chapter 18, Problem 18.152RP

(a)

To determine

The angular acceleration (α) of the component.

(b)

To determine

The dynamic reaction at A (A) and B (B) immediately after the couple is applied.

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Chapter 18, Problem 18.151RP

Chapter 18, Problem 18.153RP

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Chapter 18 Solutions

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics

Ch. 18.1 - A thin, homogeneous disk of mass m and radius r...Ch. 18.1 - Prob. 18.2P Ch. 18.1 - 18.3 Two uniform rods AB and CE, each of weight 3...Ch. 18.1 - A homogeneous disk of weight W = 6 lb rotates at...Ch. 18.1 - Prob. 18.5P Ch. 18.1 - A solid rectangular parallelepiped of mass m has a...Ch. 18.1 - Prob. 18.8P Ch. 18.1 - Determine the angular momentum HD of the disk of...Ch. 18.1 - Prob. 18.10P Ch. 18.1 - Prob. 18.11P

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The angular acceleration ( α ) of the component.

Solution Summary: The author calculates the mass per unit area of the steel using the relation: rho =mA